Could someone please explain how to solve this problem cheers.
$\displaystyle 1-x^2+x^4-x^6+\cdots=\sum_{n=0}^{\infty}(-1)^nx^{2n}$
The radius of convergence can be found using the Root test...we must have that $\displaystyle \lim_{n\to\infty}\sqrt[n]{|(-1)^nx^{2n}|}=\lim_{n\to\infty}x^2<1$
And where the series converges we can use the resmblance to the geometric series to obtain that $\displaystyle \sum_{n=0}^{\infty}(-1)^nx^{2n}=\sum_{n=0}^{\infty}\left(-x^2\right)^n=\frac{1}{1-(-x^2)}=\frac{1}{1+x^2}$
$\displaystyle 1-x^{2}+x^{4}-x^{6}+\cdots +(-1)^nx^{2n}+\cdots=\sum\limits_{n=0}^{\infty }{(-1)^{n}x^{2n}}.$
Now put $\displaystyle a_{n}=(-1)^{n}x^{2n}$ then $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{\left| a_{n} \right|}=\underset{n\to \infty }{\mathop{\lim }}\,x^{2}=x^{2}$ and the convergence requires that $\displaystyle x^2<1\implies-1<x<1.$